Simplify and expand the following expression: $ \dfrac{5}{4q + 40}+ \dfrac{4}{4q - 28}- \dfrac{3}{q^2 + 3q - 70} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{5}{4q + 40} = \dfrac{5}{4(q + 10)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{4}{4q - 28} = \dfrac{4}{4(q - 7)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 + 3q - 70} = \dfrac{3}{(q + 10)(q - 7)}$ Now we have: $ \dfrac{5}{4(q + 10)}+ \dfrac{4}{4(q - 7)}- \dfrac{3}{(q + 10)(q - 7)} $ The least common multiple of the denominators is: $ 16(q + 10)(q - 7)$ In order to get the first term over $16(q + 10)(q - 7)$ , multiply by $\dfrac{4(q - 7)}{4(q - 7)}$ $ \dfrac{5}{4(q + 10)} \times \dfrac{4(q - 7)}{4(q - 7)} = \dfrac{20(q - 7)}{16(q + 10)(q - 7)} $ In order to get the second term over $16(q + 10)(q - 7)$ , multiply by $\dfrac{4(q + 10)}{4(q + 10)}$ $ \dfrac{4}{4(q - 7)} \times \dfrac{4(q + 10)}{4(q + 10)} = \dfrac{16(q + 10)}{16(q + 10)(q - 7)} $ In order to get the third term over $16(q + 10)(q - 7)$ , multiply by $\dfrac{16}{16}$ $ \dfrac{3}{(q + 10)(q - 7)} \times \dfrac{16}{16} = \dfrac{48}{16(q + 10)(q - 7)} $ Now we have: $ \dfrac{20(q - 7)}{16(q + 10)(q - 7)} + \dfrac{16(q + 10)}{16(q + 10)(q - 7)} - \dfrac{48}{16(q + 10)(q - 7)} $ $ = \dfrac{ 20(q - 7) + 16(q + 10) - 48} {16(q + 10)(q - 7)} $ Expand: $ = \dfrac{20q - 140 + 16q + 160 - 48}{16q^2 + 48q - 1120} $ $ = \dfrac{36q - 28}{16q^2 + 48q - 1120}$ Simplify: $ = \dfrac{9q - 7}{4q^2 + 12q - 280}$